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Force - conservative or not?

  1. Mar 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A particular spring obeys the force law ## \vec F= (-kx + ax^3 +bx^4)\vec i##. Is this force conservative? Explain why or why not.

    2. Relevant equations
    Conservative forces depend on the beginning and endpoints, but not the path taken.

    3. The attempt at a solution
    The given equation for force clearly depends on ##x##, which is distance. Therefore, this force is non-conservative. Additionally, if ##\vec F## depends on ##x##, then the graph of F vs dl will vary depending on what value of ##x## is plugged in. Work (which is change in energy) will not be constant, so once again force is not conserved. My answer is incorrect, according to my solutions guide. Where did I mess up?
     
  2. jcsd
  3. Mar 27, 2015 #2

    ShayanJ

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    Gold Member

    Here's where you messed up:
    This is wrong. Conservative forces are forces that their work on a particle in moving it from one point to another, doesn't depend on the path taken and only depends on the two points. This is equivalent to saying that a force is conservative if and only if ## \oint \vec F \cdot \vec{dl}=0 ## for all closed paths. It can be proved that this is equivalent to ## \vec \nabla \times \vec F=0 ##. So the latter equation is a good way of finding out whether a force field is conservative or not.
     
  4. Mar 27, 2015 #3
    What does the symbol integral with circle mean?

    The second part of the question asks me to find the potential energy function, which is the integral of ## \vec F ##. If force is conservative, then the potential energy function is equal to zero right?
     
  5. Mar 27, 2015 #4

    ShayanJ

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    It means the line integral is taken over a closed path.
    No. To obtain the potential energy from the force field, you should use ## \phi(\vec r)=-\int_{\vec{r}_{ref}}^{\vec r} \vec F \cdot \vec{dl} ## where ## \phi(\vec r _{ref}) ## is defined to be zero, i.e. the point where potential becomes zero is arbitrary.
     
  6. Mar 27, 2015 #5
    In one dimension (as it is this case) a force ##\vec{F}## is conservative if and only if there is a function ##\phi(x)## such that ##\vec{F} = -\frac{d}{dx}\phi(x) \vec{i}##

    In your example the force is a continuous function of the position ##x## so the above is satisfied (i.e. the force is conservative).
     
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